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The e.m.f. of the cell : Cu(s)"|"Cu^(...

The e.m.f. of the cell :
`Cu(s)"|"Cu^(2+)(1M)"||"Ag^(+)(1M)"|"Ag`
is 0.46 V. The standard reduction reduction potential of `Ag^(+)//Ag` is 0.80 V. The standard reduction potential of `Cu^(2+)//Cu` is

A

`-0.34`V

B

1.26 V

C

`-1.26` V

D

`0.34` V

Text Solution

Verified by Experts

The correct Answer is:
D
`E_("cell")^(@)=E^(@)(Ag^(+)"|"Ag)-E^(@)(Cu^(2+) "|"Cu)`
`0.46 = 0.80 - E^(@)(Cu^(2+)"|"Cu)`
`:.E^(@)(Cu^(2+)"|"Cu)= 0.80-0.46= 0.34V`
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